Rewrite the function by completing the square. $f(x)=x^{2}+4x-60$ $f(x)=(x+$
Explanation: We want to complete $x^2{+4}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+4}}{2}\right)^2={4}$ to it: $x^2{+4}x+{4}=(x+2)^2$ In order to keep the expression equivalent, we add and subtract ${4}$, not forgetting the expression's constant term, $-60$ : $\begin{aligned} f(x)&=x^2+4x-60 \\\\ &=x^2+4x+{4}-60-{4} \\\\ &=(x+2)^2-60-4 \\\\ &=(x+2)^2-64 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 2)^2 - 64$